\newproblem{lay:2_1_12}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 2.1.12}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
  Let $A=\begin{pmatrix}3 & -6 \\ -2 & 4\end{pmatrix}$. Construct a $2\times 2$ matrix $B$ such that $AB$ is the zero matrix. Use two different nonzero
	columns for $B$
}{
  % Solution
	We search for a matrix $B=\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{pmatrix}$ such that
	\begin{center}
	   $AB=\begin{pmatrix}3 & -6 \\ -2 & 4\end{pmatrix}\begin{pmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{pmatrix}=
		     \begin{pmatrix}  3b_{11}-6b_{21} & 3b_{12}-6b_{22} \\ 4b_{21}-2b_{11}& 4b_{22}-2b_{12}\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix}$
	\end{center}
	This matrix equation gives us 4 equations
	\begin{center}
	   $3b_{11}-6b_{21}=0$ \\
		 $3b_{12}-6b_{22}=0$ \\
		 $4b_{21}-2b_{11}=0$ \\
		 $4b_{22}-2b_{12}=0$
	\end{center}
	The augmented matrix of this equation system is
	\begin{center}
	   $\left(\begin{array}{rrrr|r} 3 & 0 & -6 & 0 & 0 \\ 0 & 3 & 0 & -6 & 0 \\ -2 & 0 & 4 & 0 & 0 \\ 0 & -2 & 0 & 4 & 0 \end{array}\right) \sim 
		  \left(\begin{array}{rrrr|r} 1 & 0 & -2 & 0 & 0 \\ 0 & 1 & 0 & -2 & 0 \\  0 & 0 & 0 & 0 & 0 \\ 0 &  0 & 0 & 0 & 0 \end{array}\right)$
	\end{center}
	Consequently, $b_{11}=2b_{21}$ and $b_{12}=2b_{22}$. That is, any matrix of the form
	\begin{center}
		$B=\begin{pmatrix}2b_{21} & 2b_{22} \\ b_{21} & b_{22}\end{pmatrix}$
	\end{center}
	yields $AB=0$. One such example is $B=\begin{pmatrix}2 & 2 \\ 1 & 1\end{pmatrix}$
}
\useproblem{lay:2_1_12}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
